Mathematics, 15.04.2020 23:16 jeanieb

I need solution and answer

1. scott invested $1600 into a retirement account that earns 2.4% interest compounded monthly. what will be the balance of the account after 30 years?

2. kaylee used her graduation money to set up a savings account that earns 3.4% interest compounded weekly. if the original amount deposited was $500 how much interest will she have earned after 10 years?

3. mr and mrs rainer took out a $240,000 loan to purchase their home. if the interest rate on the loan is 1.2% compounded bimonthly , how much interest will they have paid after 30 years?

Answers: 3

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Answer from: jamaicalove2880

Using compound interest, it i found that:

1. The balance will be of $3,284.74.

2. She will have earned $202.4 in interest after 10 years.

3. They will have paid $103,968 in interest after 30 years.

Compound interest:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

A(t) is the amount of money after t years. P is the principal(the initial sum of money). r is the interest rate(as a decimal value). n is the number of times that interest is compounded per year. t is the time in years for which the money is invested or borrowed.

Item 1:

Invested $1600, hence P = 1600

Rate of 2.4%, hence r = 0.024

Compounded monthly, hence n = 12

30 years, hence t = 30

The balance is:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

$A(30) = 1600\left(1 + \frac{0.024}{12}\right)^{30(12)}$

A(30) = 3284.73

The balance will be of $3,284.74.

Item 2:

Invested $500, hence P = 500

Rate of 3.4%, hence r = 0.034

Compounded weekly, hence n = 52

10 years, hence t = 10

The balance is:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

$A(10) = 500\left(1 + \frac{0.034}{52}\right)^{52(10)}$

A(10) = 702.4

Interest is balance subtracted by principal, hence, she will have earned $202.4 in interest after 10 years.

Item 3:

Loan of $240000, hence P = 240000

Rate of 1.2%, hence r = 0.012

Compounded bimonthly, hence n = 24

30 years, hence t = 30

The balance is:

$A(t) = P\left(1 + \frac{r}{n}\right)^{nt}$

$A(30) = 240000\left(1 + \frac{0.012}{24}\right)^{24(30)}$

A(30) = 343968

Interest is balance subtracted by principal, hence:

343968 - 240000 = 103,968

They will have paid $103,968 in interest after 30 years.

A similar problem is given at link

Answer from: computer15

Step-by-step explanation:

1. A = P (1 + r/n)^(nt)

A = 1600 (1 + 0.024/12)^(12×30)

A = 3284.73

2. A = P (1 + r/n)^(nt)

A = 500 (1 + 0.034/52)^(52×10)

A = 702.40

3. A = P (1 + r/n)^(nt)

A = 240,000 (1 + 0.012/6)^(6×30)

A = 343,875.41

A − P = 103,875.41

Answer from: Quest

answer 1 : option "c"

answer 2 : option "d"

answer 3 : option "b"

answer 4 : option "d"

answer 5 : option "b"

answer 6 : option "a"

answer 7 : option "c"

answer 8 : option "d"

answer 9 : option "c"

answer 10 : option "d"

answer 11 : option "c"

answer 12 option "b"

answer 13 option "c"

answer 14 option "c"

answer 15 option "d"

answer 16 option "a"

answer 17 option "b"

answer 18 option "c"

answer 19 option "a"

answer 20 option "a"

answer 21 option "c"

Answer from: Quest

step-by-step explanation:

Someone and explain as well as answer the question u

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